CH 302 Unit 2 Review 1 INTRODUCTION TO CHEMICAL EQUILIBRIUM

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Jemimah Bradford
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1 CH 302 Unit 2 Review 1 INTRODUCTION TO CHEMICAL EQUILIBRIUM

2 IntroducBon to Chemical Equilibrium A working defini-on for equilibrium is the state of a chemical reac-on when the rate of the forward reac-on and the reverse reac-on are equal. At this point, there is no net change in the concentra-ons of your reac-on We use the equilibrium constant, K, to calculate these exact amounts at equilibrium: aa + bb # eq cc + dd K = a C c a D da a Aa a B b Mathema-cally, K is equal to the ra-o of the ac-on of the products raised to the power of their coefficients divided by the ac-on of the reactants raised to the power of their coefficients. This is Mass Ac-on.

aa + bb # eq cc + dd IntroducBon to Chemical Equilibrium We can directly correlate ac-on to pressure and concentra-on to create a more sensible rela-onship.

3 aa + bb # eq cc + dd IntroducBon to Chemical Equilibrium We can directly correlate ac-on to pressure and concentra-on to create a more sensible rela-onship. K = a C c a D da a Aa a B b a i = [i] [i] a i = P i P i REMEMBER: Assume for this example that all species in the K c example are aqueous. All the species in the K p example are gases. Remember: the ac-on of any LIQUID or SOLID is 1. These terms will drop out of the mass ac-on expression. K c = [C]c [D] da [A] a [B] b K = P c P da C D p P a P b A B

4 Easy QuesBon Write the mass ac-on expression in terms of K c for the following reac-on: eq Ag PO (s) # 3Ag + (aq)+ PO 3 (aq) Follow- up Ques-on: What term from Unit 1 is this equal to?

6 Challenging QuesBon Write the mass ac-on expression in terms of K c for the following reac-on: eq Al(NO 3 ) 3 (aq)+ 3NaOH(aq) # Al(OH ) 3 (s)+ 3NaNO 3 (aq)

8 Q vs K: Chemical Equilibrium Terminology The purpose of K is to tell you the concentra-ons at equilibrium. The purpose of Q is to tell you the concentra-ons at any given star-ng point. The purpose of comparing Q to K is to tell you how the reac-on will proceed at your Q- concentra-ons. 1. Q < K : reac-on moves forward toward equilibrium 2. Q = K : reac-on is at equilibrium (just right) 3. Q > K : reac-on moves backward toward equilibrium

$Free Energy vs K When K is small (less than 1), only a small frac-on of your reactants becomes products.$ When K is large (greater than 1), a greater amount of products are formed than reactants remain.

When K is large (greater than 1), a greater amount of products are formed than reactants remain.

9 Free Energy vs K When K is small (less than 1), only a small frac-on of your reactants becomes products. A reac-on with a small K reacts to a small extent. When the reactant is favored, the ΔG is posi-ve. When K is large (greater than 1), a greater amount of products are formed than reactants remain. A reac-on with a large K reacts to a large extent. When the product is favored, the ΔG is nega-ve. Note: these graphs are an exaggera-on of K values. Very large K (think combus-on) and very small K (think solubility examples) do not really look like this on a graph

10 Visualizing Free Energy, K, and Q (Checklist) ΔG < 0 Iden-fy the rela-onship between Q and K at any given point on the graph Determine whether the reac-on is spontaneous or non- spontaneous Is ΔG posi-ve or nega-ve for the reac-on? Is K greater than or less than 1 on the graph? Is ΔG posi-ve or nega-ve at any given point? Advanced: How will the graph change if you compared to another reac-on with a ΔG of greater magnitude? How will the graph change if you increase K?

11 Visualizing Free Energy, K, and Q ΔG < 0 A. Q is equal to 0. No mader what K is, Q will be less than K. Reac-on will move forward toward equilibrium. B. Q is less than K. Reac-on will move forward. C. Q is equal to K. Reac-on is at equilibrium. D. Q is greater than K. Reac-on will slope back toward the reactants. E. Q is infinity. No mader what K is, Q will be greater than K. Reac-on will move backward toward equilibrium.

12 QuanBfying Free Energy, K, Q Free Energy, K, and Q are all related based on the following formulas: ΔG r = ΔG r + RT lnq ΔG r = RT ln K K = e ΔG r RT The free energy of a reac-on under any measurable ini-al condi-ons At equilibrium, ΔG r = 0. Also, Q = K. Therefore, we get a new equa-on for the rela-onship between standard free energy and K. We can rearrange this equa-on to solve directly for a K value at a given temperature given the standard free energy change.

13 CalculaBon For the following reac-on A(aq) B(aq) + C(aq) The value of K c = 1.8 x What is the ΔG for this reac-on at 298K?

15 ApplicaBon QuesBon Consider the reac-on: A(g) 2B(g) 3atm of A is added to an empty chamber. Once the reac-on reaches equilibrium, there is 5atm in total. What is the K p for this reac-on?

17 Final QuesBon (HW QuesBon 15) The system H 2 (g) + I 2 (g) 2HI(g) is at equilibrium at a fixed temperature with a par-al pressure of H 2 of atm, a par-al pressure of I 2 of atm, and a par-al pressure of HI of atm. An addi-onal 0.26 atm pressure of HI is admided to the container, and it is allowed to come to equilibrium again. What is the new par-al pressure of HI?

20 Temperature Dependence of K Once incredibly important (and some-mes overlooked) rela-onship is K and Temperature. K s dependence on temperature depends on whether the reac-on is endothermic or exothermic. The van t Hoff Equa-on is: ln( K 2 K 1 ) = ΔH rxn R ( 1 T 1 1 T 2 ) When you do the math, you will see that for endothermic reac-ons, increasing temperature increases K. For exothermic reac-ons, increasing temperature decreases K. It all depends on the sign of ΔH rxn Play with some values to prove this rela-onship, then next week I will give you an easier way to figure this out.

CH 302 Exam Review CHEMICAL EQUILIBRIUM, ACIDS & BASES THINKING LIKE A CHEMIST

CH 302 Exam Review CHEMICAL EQUILIBRIUM, ACIDS & BASES THINKING LIKE A CHEMIST Exam Two Learning Outcomes Learning Outcomes for Chemical Equilibrium Describe the rela-onship between free energy and equilibrium